2.02 Function notation

We have seen that functions are special types of relations. When dealing with functions we often use function notation to emphasise that we are dealing with a special case, to highlight the independent and dependent variable and for ease of notation with substitution and later in the year for calculus.

When we are writing in function notation, instead of writing "$y=$y=", we write "$f(x)=$f(x)=". This can be interpreted as "$f$f is a function of the variable $x$x" and read as "$f$f of $x$x". Common letters to use for general functions are lower case $f$f, $g$g and $h$h. However, any letter can be used and we can use variables that have meaning in context. 

Exploration

Instead of $y=2x+1$y=2x+1 we could write $f(x)=2x+1$f(x)=2x+1, here $f$f is a function of the variable $x$x which follows the rule double $x$x and add $1$1.

$P(t)=200\times0.8^t$P(t)=200×0.8t, here population is a function of time.

$H(d)=30-2d-30d^2$H(d)=30−2d−30d2, here height is a function of distance.

These are all examples of just a different way to write '$y=$y= …' notation as a function. The letter in the bracket on the left is the input and the right hand side gives us a rule for the output.

Function notation also allows for shorthand for substitution. If $y=3x+2$y=3x+2 and we write this in function notation as $f(x)=3x+2$f(x)=3x+2, then the question 'what is the value of $y$y when $x$x is $5$5?' can be asked simply as 'what is $f(5)$f(5)?' or 'Evaluate $f(5)$f(5).'

 

Worked example

If $A(x)=x^2+1$A(x)=x2+1 and $Q(x)=x^2+9x$Q(x)=x2+9x, find:

(a) $A(5)$A(5)

Think: This means we need to substitute $5$5 in for $x$x in the $A(x)$A(x) equation.

Do: 

$A(5)$A(5)	$=$=	$5^2+1$52+1
	$=$=	$26$26

 

(b) $Q(6)$Q(6)

Think: This means we need to substitute $6$6 in for $x$x in the $Q(x)$Q(x) equation.

Do:

$Q(6)$Q(6)	$=$=	$6^2+9\times6$62+9×6
	$=$=	$36+54$36+54
	$=$=	$90$90

 

Practice questions

Question 1

Question 2

 

Algebra of functions

Functions can be added, subtracted, multiplied or divided. If $f$f and $g$g are functions, we compute $f(x)+g(x)$f(x)+g(x) by adding the function values at $x$x for each function. The short-hand notation for function addition is $(f+g)(x).$(f+g)(x). In a similar way, we form $f(x)-g(x)$f(x)−g(x) or $(f-g)(x)$(f−g)(x) for subtraction, $f(x)\times g(x)$f(x)×g(x) or $(fg)(x)$(fg)(x) for multiplication and $f(x)\div g(x)$f(x)÷​g(x) or $(f/g)(x)$(f/g)(x) for function division.

 

Practice questions

Question 3

question 4

Composition of functions

The idea behind the composition of functions is best explained with an example.

Suppose we think about the function given by $f\left(x\right)=2x+1$f(x)=2x+1. We understand that the function takes values of $x$x for the input and maps them to values, say $y=2x+1$y=2x+1, in the output.

Suppose however that this is only the first part of a two-stage treatment of $x$x. Suppose we now take these function values and map them using another function, say $g\left(x\right)=x^2$g(x)=x2. This means that the $y$y values given by $\left(2x+1\right)$(2x+1) become the squared values $\left(2x+1\right)^2$(2x+1)2. The diagram below captures the idea.

The function values $f\left(x\right)$f(x) have become the input for $g\left(x\right)$g(x). Thus we could describe the complete two-stage process by the expression $g\left(f\left(x\right)\right)$g(f(x)), sometimes written $(g\circ f)(x)$(g∘f)(x) and spoken of as the "$g$g of $f$f  of $x$x".

Algebraically, we can write $g\left(f\left(x\right)\right)=g\left(2x+1\right)=\left(2x+1\right)^2$g(f(x))=g(2x+1)=(2x+1)2. 

Note that if we reversed the order of the two-stage processing, we would, in this instance, develop a different composite function. 

Thus $f\left(g\left(x\right)\right)=f\left(x^2\right)=2\left(x^2\right)+1=2x^2+1$f(g(x))=f(x2)=2(x2)+1=2x2+1. This would be "$f$fof $g$g of $x$x".

 

Practice questions

Question 5

Question 6